Find a sequence of consecutive integers, a, a+1, a+2, ..., b such that b is a square and that:
[a+(a+1)+(a+2)+...+(b-1)]b and [(a-1)+a+(a+1)+(a+2)+...+(b-1)]b are both 10-digit numbers containing all of the digits 0 to 9.
Find closed forms for expressions 1 and 2 in terms of a and b.
expression1 = b*((b+a-1)*(b-a)/2)
expression2 = b*((b+a-2)*(b-a+1)/2) = expression1 + (a-1)b
Since b is a square, find range of possible integer values for √b
First, a can be as small as 1 and as large as b-1.
Since both expressions must be 10 digits, this limits √b to 35 to 316
Result: I found only one solution for √b from 35 ... 316 inclusive
a b √b expression1 expression2
78 1296 36 [1083659472, 1083759264]
The requested sequence is:
(a, ..., b) = (78,79,80, ... 1274,1275,1296)
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def issquare(n):
""" Input an integer, Returns True iff it is a perfect square. """
if round(n**0.5)**2 == n:
return True
else:
return False
def s(a,b):
if a >= b:
return 0
x = b*int((b+a-1)*(b-a)/2)
return [x, x + (a-1)*b]
from itertools import permutations
pans = []
for perm in permutations('0123456789'):
if perm[0] == '0':
continue
pans.append(int(''.join(perm)))
for sqrtb in range(35, 317):
print(sqrtb)
b = sqrtb**2
for a in range(1,b):
ns = s(a,b)
if len(str(ns[0])) != 10:
continue
if len(str(ns[1])) != 10:
continue
if ns[0] not in pans:
continue
if ns[1] not in pans:
continue
print(a,b,sqrtb,s(a,b))
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Posted by Larry
on 2023-12-25 09:45:52 |
Solution
